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In most earth observation satellite specifications, the Panchromatic (PAN) Ground Sampling Distance (GSD) is 4 times the Multi-spectral (MS) GSD. But, I cannot find a detailed explanation on why that is.

If a PAN color detector pixel size is 1μm, I think the MS (4 band) mono detector pixel size will be 2μm due to the Bayer filter.

If this is correct, the PAN GSD should be 2 times the MS GSD, not 4.

Am I doing something wrong when determining the GSD for these two sensor types?

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It depends on the satellite, used type of sensor and so on.

Most satellites don't have cameras with bayer sensors (citation needed) but monochrome sensors with filters. Often those filters are mounted on wheels that can rotate one of a whole set of filters for different wavelengts in front of the sensor. The benefit of that is, that you have full resolution for the image in the given spectrum and that you can have (many) more than three filters on a wheel (the only limit is the size of the wheel). So you can have different IR and UV filters, wideband, narrowband, stopband filters with one camera. The downside is, that you only get to use one filter at a time which can lead to artifacts in the image when imaging moving stuff...

Satellites in low earth orbit often have line scan cameras on board that scan the earth like your flatbed scanner while moving relative to the surface of earth. Sometimes several lines are used with different filters.

What every filter does is removing part of the light that hits the sensor. So the image get's darker. To counteract this we either have to increase the exposure time (not always an option), open the aperture (not always possible) or increase the pixel gain (also described as "sensitivity / ISO" on consumer cameras, which is not technically correct). By increasing the gain we also increase noise. One method to reduce noise is by binning senso picels together, basically adding the values of 2x2 or 3x3 pixels together. Many scientific/industrial cameras can do this in hardware already so I can imagine that the imagers on satellites can do this as well. Otherwise it can be done in software too (with possible slight drawbacks in very dark image regions).

With a line scanning image sensor you can increase the exposure time (which will reduce the resolution of the image in the x direction) or do binning (reducing the resolution the the y direction) or both of them together (there is no law stating that x and y spatial resolution must always be the same)...

So it could be that they use different (more sensitive with bigger pixels) sensors for multispectral images. Or that they use binning. Or that they just want/need to save bandwith when downlinking the images.

I guess it's not always the same and sometimes a combination of reasons.

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Simple math shows that a 2μm square pixel is four times the area of a 1μm square pixel.

Therefore, the PAN GSD (using a 2μm square pixel) resolution would be 4 times the MS GSD (using a 1μm square pixel) resolution.

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    \$\begingroup\$ I'm not sure if the area comes into play. The wiki on GSD states: "ground sample distance (GSD) [...] is the distance between pixel centers measured on the ground". Using that definition, the GSD should scale linearly with the pixel size, right? Perhaps there exist multiple definitions of GSD? \$\endgroup\$ Commented Mar 31, 2022 at 12:35
  • \$\begingroup\$ @SaaruLindestøkke The sample distance doubles horizontal and doubles vertically. Ordinary language doesn’t handle this sort of thing well. That’s even a problem for people who work with these things professionally sometimes when it’s not directly a matter of a paycheck. \$\endgroup\$ Commented Mar 31, 2022 at 20:52
  • \$\begingroup\$ Well, I get that if the GSD was defined on two axes that it would "double twice". Perhaps that is what's meant with GSD resolution? That it's in two dimensions? However, when using a simple GSD calculator (for drones) I find that going from a (fictional) sensor with 1 micrometer pixels to a sensor with 2 micrometer pixels doubles the GSD from 0.2 to 0.4 (as the question asker found themselves as well). I'm not an expert, but maybe terms are being mixed up? \$\endgroup\$ Commented Mar 31, 2022 at 22:20

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